Question

If the equation of the plane passing through point $ (3, 2, 5) $ and perpendicular to the planes $$ 2x - 3y + 5z = 7, \quad 5x + 2y - 3z = 11 $$ is $$ x + by + cz + d = 0, $$ then find $ 2b + 3c + d $.

• A. 0
• B. 35
• C. 1
• D. 20

Hint:

Cross product of normal vectors gives normal to plane perpendicular to both.

Answer 1

The Correct Option is B

Normal vector to required plane is cross product of normals of given planes: \[ \vec{n_1} = (2, -3, 5), \quad \vec{n_2} = (5, 2, -3) \] \[ \vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 5 \\ 5 & 2 & -3 \\ \end{vmatrix} = ( (-3)(-3) - 5(2), \; 5(5) - 2(-3), \; 2(2) - (-3)(5)) = (9 - 10, 25 + 6, 4 + 15) = (-1, 31, 19) \] Equation of plane: \[ -1(x - 3) + 31(y - 2) + 19(z - 5) = 0 \] \[ - x + 3 + 31y - 62 + 19z - 95 = 0 \Rightarrow -x + 31y + 19z - 154 = 0 \] Rewrite as: \[ x - 31y - 19z + 154 = 0 \] Compare with \( x + b y + c z + d = 0 \), so: \[ b = -31, \quad c = -19, \quad d = 154 \] Calculate: \[ 2b + 3c + d = 2(-31) + 3(-19) + 154 = -62 - 57 + 154 = 35 \]